Proof of "Freshman's dream" in commutative rings 
When $p$ is a prime number and $x$ and $y$ are members of a
  commutative ring of characteristic $p$, then $$(x+y)^p=x^p+y^p.$$ 
  This
  can be seen by examining the prime factors of the binomial
  coefficients: the $n$th binomial coefficient is
$$\binom{p}{n} = \frac{p!}{n!(p-n)!}.$$
The numerator is $p$ factorial, which is divisible by $p$. However,
  when $0 < n < p$, neither $n!$ nor $(p - n)!$ is divisible by $p$
  since all the terms are less than $p$ and $p$ is prime. Since a
  binomial coefficient is always an integer, the $n$th binomial
  coefficient is divisible by $p$ and hence equal to 0 in the ring. We
  are left with the zeroth and $p$th coefficients, which both equal 1,
  yielding the desired equation.

1) My immediate thought is, so when we say binomial coefficient we mean addition forms also: for example, $3x^2y = x^2y+x^2y+x^2y$. But can this be generalized to all rings of characteristic $p$?
2) Also, how is coefficient being divided by $p$ related to coefficient becoming zero in characteristic $p$?
 A: This should answer both of your questions.
Characteristic $p > 0$ means (informally, see the post by Thomas Andrews for a formal definition) that the multiple
$$
p \cdot 1 = \underbrace{1+ \dots + 1}_{\text{$p$ times}}
$$
is zero. It follows $p \cdot a = 0$ for all $a$ in the ring. This is because $$p \cdot a = \underbrace{a + \dots + a}_{\text{$p$ times}} = (\underbrace{1+ \dots + 1}_{\text{$p$ times}}) a = (p \cdot 1) a = 0 a = 0.$$
Also, if $p$ divides the integer $n$, so that $n = m p$ for some $m$, then $$n \cdot a = (m p) \cdot a = m \cdot (p \cdot a) = m \cdot 0 = 0.$$
Now the binomial theorem in a commutative ring is
$$
(x + y)^{n} = \sum_{i=0}^{n} \binom{n}{i} \cdot x^{i} y^{n-i},
$$
where note that the binomal coefficients act as multiples.
A: 1) You need to deal with the order of multiplication in non-commutative rings, so it's no longer $3x^2y$ in the binomial theorem but $xxy + xyx + yxx$. I'm not entirely sure if it breaks down in this case, but it seems like it would.
2) If the coefficient is divisible by $p$, then it is equal to $pq$ for some $q$. If the ring has characteristic $p$, then $pq$ = $0q$ = $0$.
